# Question #56893

Mar 18, 2017

$\left(\begin{matrix}a & b & c \\ 1 & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ 1 & - \frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{2} \\ - 1 & \frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{2} \\ - 1 & - \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{matrix}\right)$

#### Explanation:

Making $a = A , b = \frac{B}{\sqrt{2}} , c = \frac{C}{\sqrt{2}}$ and substituting we have

${A}^{4} + {B}^{4} + {C}^{4} - 4 A B C = - 1$

Making $A = B = C = X$ (because of symmetry) we have

$3 {X}^{4} - 4 {X}^{3} + 1 = 0$

by inspection we have the solutions

$X = 1 , X = 1 , X = \frac{1}{3} \left(- 1 \pm i \sqrt{2}\right)$ so the real solutions are

$A = B = C = 1$ or

$a = 1 , b = \frac{\sqrt{2}}{2} , c = \frac{\sqrt{2}}{2}$

There are four symmetries so there are four solutions

$\left(\begin{matrix}a & b & c \\ 1 & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ 1 & - \frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{2} \\ - 1 & \frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{2} \\ - 1 & - \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{matrix}\right)$